How to Solve Civil Service Exam Number Series Problems 2

In the previous post, we have learned how to solve number sequence (for the Civil Service Exam Number Series test) and letter sequence problems that involves constant difference or constant skips. In this post, we are going to discuss another type of sequence. Before we discuss, see if you can find the next term of the following sequences.

1. 3, 6, 12, 24, 48, ___

2. 18, 6, 2, 0.66…, ___

3. 1, 4, 9, 16, 25, ___

4. 3, 12, 27, 48, 75, ___

Solution and Explanation

First Sequence: 3, 6, 12, 24, 48, ___

In the first sequence, the first that you will notice is that the second term is twice the first term. So, the next thing that you should ask is, “Is the third term twice the second term?” Yes, 12 is twice 6. What about the next term? Yes. So, each term in the sequence is multiplied to 2 to get the next term. Therefore, the missing term is 96 which is 48 multiplied by 2.

If we look at the difference of the numbers in the sequence above, we can see that the number we add is also increasing twice. To get 6, we added 3. To get 12, we added 6. To get 24, we add 12 and so on. As we can see, the sequence of the numbers we add (the numbers in red color) is the same as the original sequence (numbers in blue color).

Second Sequence: 18, 6, 2, 0.66…, ___

In the second sequence, the number is reduced each time. Since they are integers, it can either be subtraction or division. As we can see, 6 is a third of 18. This means that to get 6, 18 is divided by 3. Now, look at the next term. It’s 2. So it is also a third of 6. Can you see the pattern now?

Each term is divided by 3 to get the next term. So, we must divide 0.66… by 3. therefore, the next term is 0.22… The three dots means that the 2’s are infinitely many.

Third Sequence: 1, 4, 9, 16, 25, ___

What is familiar with this sequence? They are all square numbers! That is,

, , , and .

So the next term is which is 36.

Fourth Sequence: 3, 12, 27, 48, 75, ___

The fourth sequence seems difficult, but I have just multiplied each number in the third sequence by 3. So, if the sequence is not familiar, try to see if you can divide it by any number. As you can see,